Mixing
Proving nicely that $3+omega=omega$ where $omega$ is the smallest limit ordinal.
0
$begingroup$
I just learned about ordinal numbers and I am a bit confused to show things with them. Let $omega$ be the smallest limit ordinal.
I want to show that $3+omega=omega$. So the definition I have of $3+omega$ (since $omega$ is a limit ordinal) is :
$3+omega = sup{n+3:n<omega }$. Then the definition of the sup I have is $supX={gamma : exists x in X text{such that } gamma in x}$, so we have :
$3+omega = {gamma : exists x in {n+3:n<omega } text{such that } gamma in x}$. But how to show this is equal to $omega$?
ordinals
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
$endgroup$
add a comment |
0
$begingroup$
I just learned about ordinal numbers and I am a bit confused to show things with them. Let $omega$ be the smallest limit ordinal.
I want to show that $3+omega=omega$. So the definition I have of $3+omega$ (since $omega$ is a limit ordinal) is :
$3+omega = sup{n+3:n<omega }$. Then the definition of the sup I have is $supX={gamma : exists x in X text{such that } gamma in x}$, so we have :
$3+omega = {gamma : exists x in {n+3:n<omega } text{such that } gamma in x}$. But how to show this is equal to $omega$?
ordinals
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
$endgroup$
1
$begingroup$
Incidentally, "limit ordinal" is often defined to include $emptyset$, making $omega$ the second smallest.
$endgroup$
– J.G.
Jan 22 at 13:04
add a comment |
0
0
0
$begingroup$
I just learned about ordinal numbers and I am a bit confused to show things with them. Let $omega$ be the smallest limit ordinal.
I want to show that $3+omega=omega$. So the definition I have of $3+omega$ (since $omega$ is a limit ordinal) is :
$3+omega = sup{n+3:n<omega }$. Then the definition of the sup I have is $supX={gamma : exists x in X text{such that } gamma in x}$, so we have :
$3+omega = {gamma : exists x in {n+3:n<omega } text{such that } gamma in x}$. But how to show this is equal to $omega$?
ordinals
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
$endgroup$
I just learned about ordinal numbers and I am a bit confused to show things with them. Let $omega$ be the smallest limit ordinal.
I want to show that $3+omega=omega$. So the definition I have of $3+omega$ (since $omega$ is a limit ordinal) is :
$3+omega = sup{n+3:n<omega }$. Then the definition of the sup I have is $supX={gamma : exists x in X text{such that } gamma in x}$, so we have :
$3+omega = {gamma : exists x in {n+3:n<omega } text{such that } gamma in x}$. But how to show this is equal to $omega$?
ordinals
ordinals
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
asked Jan 22 at 13:03
roi_saumonroi_saumon
59438
59438
1
$begingroup$
Incidentally, "limit ordinal" is often defined to include $emptyset$, making $omega$ the second smallest.
$endgroup$
– J.G.
Jan 22 at 13:04
add a comment |
1
$begingroup$
Incidentally, "limit ordinal" is often defined to include $emptyset$, making $omega$ the second smallest.
$endgroup$
– J.G.
Jan 22 at 13:04
1
1
$begingroup$
Incidentally, "limit ordinal" is often defined to include $emptyset$, making $omega$ the second smallest.
$endgroup$
– J.G.
Jan 22 at 13:04
$begingroup$
Incidentally, "limit ordinal" is often defined to include $emptyset$, making $omega$ the second smallest.
$endgroup$
– J.G.
Jan 22 at 13:04
add a comment |
1 Answer
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For any $n < omega$, $3+n < omega$, so $3+omega leq omega$. On the other hand, for any $n < omega$, $3+omega geq 3+n > n$. Thus $3+omega geq omega$.
answered Jan 22 at 13:07
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Tank you for the explanation. Some questions that might sound stupid : If $n < omega$ what argument can we use to say that $3+n < omega$? also you say that $n<omega$ implies $n<omega$ and thus $3+omega leq omega$. So I guess the first statement should be $nleq omega$ implies $n leqomega$. This all may seem obvious to you but a lot of things that seemed obvious to me for normal numbers are not true anymore for ordinals.
$endgroup$
– roi_saumon
Jan 22 at 13:24
$begingroup$
@roi_saumon: I listed a lot of basic ordinal arithmetic results in this 29 October 2006 sci.math post, such as: "[$+$ is] associative, not commutative, continuous in the right variable, not continuous in the left variable $(sup{n+omega: n< omega }= omega< omega cdot 2 = sup{n: n < omega } + omega ),$ strictly increasing in the right variable, non-decreasing in the left variable (not strict: $1 + omega = 2 + omega).$" The proofs of these will depend on how your book/notes define ordinal addition, multiplication, etc.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:52
$begingroup$
@DaveL.Renfro I see. I am following notes that don't have proofs for this. Any good book with these proofs that you recommend?
$endgroup$
– roi_saumon
Jan 22 at 14:55
1
$begingroup$
@roi_saumon: Maybe Basic Set Theory by Shen/Vereshchagin, which doesn't have lengthy treatments of axiomatics and other issues that are rather tangential to someone only interested in ordinal arithmetic. Possibly the most detailed treatment of ordinal arithmetic I know of in a beginning set theory text is Introduction to Set Theory by James Donald Monk (1969), if you can find it in a library. And of course there is the bible.
$endgroup$
– Dave L. Renfro
Jan 22 at 15:31
add a comment |
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1 Answer
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$begingroup$
For any $n < omega$, $3+n < omega$, so $3+omega leq omega$. On the other hand, for any $n < omega$, $3+omega geq 3+n > n$. Thus $3+omega geq omega$.
answered Jan 22 at 13:07
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Tank you for the explanation. Some questions that might sound stupid : If $n < omega$ what argument can we use to say that $3+n < omega$? also you say that $n<omega$ implies $n<omega$ and thus $3+omega leq omega$. So I guess the first statement should be $nleq omega$ implies $n leqomega$. This all may seem obvious to you but a lot of things that seemed obvious to me for normal numbers are not true anymore for ordinals.
$endgroup$
– roi_saumon
Jan 22 at 13:24
$begingroup$
@roi_saumon: I listed a lot of basic ordinal arithmetic results in this 29 October 2006 sci.math post, such as: "[$+$ is] associative, not commutative, continuous in the right variable, not continuous in the left variable $(sup{n+omega: n< omega }= omega< omega cdot 2 = sup{n: n < omega } + omega ),$ strictly increasing in the right variable, non-decreasing in the left variable (not strict: $1 + omega = 2 + omega).$" The proofs of these will depend on how your book/notes define ordinal addition, multiplication, etc.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:52
$begingroup$
@DaveL.Renfro I see. I am following notes that don't have proofs for this. Any good book with these proofs that you recommend?
$endgroup$
– roi_saumon
Jan 22 at 14:55
1
$begingroup$
@roi_saumon: Maybe Basic Set Theory by Shen/Vereshchagin, which doesn't have lengthy treatments of axiomatics and other issues that are rather tangential to someone only interested in ordinal arithmetic. Possibly the most detailed treatment of ordinal arithmetic I know of in a beginning set theory text is Introduction to Set Theory by James Donald Monk (1969), if you can find it in a library. And of course there is the bible.
$endgroup$
– Dave L. Renfro
Jan 22 at 15:31
add a comment |
1
$begingroup$
For any $n < omega$, $3+n < omega$, so $3+omega leq omega$. On the other hand, for any $n < omega$, $3+omega geq 3+n > n$. Thus $3+omega geq omega$.
answered Jan 22 at 13:07
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Tank you for the explanation. Some questions that might sound stupid : If $n < omega$ what argument can we use to say that $3+n < omega$? also you say that $n<omega$ implies $n<omega$ and thus $3+omega leq omega$. So I guess the first statement should be $nleq omega$ implies $n leqomega$. This all may seem obvious to you but a lot of things that seemed obvious to me for normal numbers are not true anymore for ordinals.
$endgroup$
– roi_saumon
Jan 22 at 13:24
$begingroup$
@roi_saumon: I listed a lot of basic ordinal arithmetic results in this 29 October 2006 sci.math post, such as: "[$+$ is] associative, not commutative, continuous in the right variable, not continuous in the left variable $(sup{n+omega: n< omega }= omega< omega cdot 2 = sup{n: n < omega } + omega ),$ strictly increasing in the right variable, non-decreasing in the left variable (not strict: $1 + omega = 2 + omega).$" The proofs of these will depend on how your book/notes define ordinal addition, multiplication, etc.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:52
$begingroup$
@DaveL.Renfro I see. I am following notes that don't have proofs for this. Any good book with these proofs that you recommend?
$endgroup$
– roi_saumon
Jan 22 at 14:55
1
$begingroup$
@roi_saumon: Maybe Basic Set Theory by Shen/Vereshchagin, which doesn't have lengthy treatments of axiomatics and other issues that are rather tangential to someone only interested in ordinal arithmetic. Possibly the most detailed treatment of ordinal arithmetic I know of in a beginning set theory text is Introduction to Set Theory by James Donald Monk (1969), if you can find it in a library. And of course there is the bible.
$endgroup$
– Dave L. Renfro
Jan 22 at 15:31
add a comment |
1
1
1
$begingroup$
For any $n < omega$, $3+n < omega$, so $3+omega leq omega$. On the other hand, for any $n < omega$, $3+omega geq 3+n > n$. Thus $3+omega geq omega$.
answered Jan 22 at 13:07
MindlackMindlack
4,885211
$endgroup$
For any $n < omega$, $3+n < omega$, so $3+omega leq omega$. On the other hand, for any $n < omega$, $3+omega geq 3+n > n$. Thus $3+omega geq omega$.
answered Jan 22 at 13:07
MindlackMindlack
4,885211
answered Jan 22 at 13:07
MindlackMindlack
4,885211
answered Jan 22 at 13:07
MindlackMindlack
4,885211
answered Jan 22 at 13:07
MindlackMindlack
4,885211
4,885211
$begingroup$
Tank you for the explanation. Some questions that might sound stupid : If $n < omega$ what argument can we use to say that $3+n < omega$? also you say that $n<omega$ implies $n<omega$ and thus $3+omega leq omega$. So I guess the first statement should be $nleq omega$ implies $n leqomega$. This all may seem obvious to you but a lot of things that seemed obvious to me for normal numbers are not true anymore for ordinals.
$endgroup$
– roi_saumon
Jan 22 at 13:24
$begingroup$
@roi_saumon: I listed a lot of basic ordinal arithmetic results in this 29 October 2006 sci.math post, such as: "[$+$ is] associative, not commutative, continuous in the right variable, not continuous in the left variable $(sup{n+omega: n< omega }= omega< omega cdot 2 = sup{n: n < omega } + omega ),$ strictly increasing in the right variable, non-decreasing in the left variable (not strict: $1 + omega = 2 + omega).$" The proofs of these will depend on how your book/notes define ordinal addition, multiplication, etc.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:52
$begingroup$
@DaveL.Renfro I see. I am following notes that don't have proofs for this. Any good book with these proofs that you recommend?
$endgroup$
– roi_saumon
Jan 22 at 14:55
1
$begingroup$
@roi_saumon: Maybe Basic Set Theory by Shen/Vereshchagin, which doesn't have lengthy treatments of axiomatics and other issues that are rather tangential to someone only interested in ordinal arithmetic. Possibly the most detailed treatment of ordinal arithmetic I know of in a beginning set theory text is Introduction to Set Theory by James Donald Monk (1969), if you can find it in a library. And of course there is the bible.
$endgroup$
– Dave L. Renfro
Jan 22 at 15:31
add a comment |
$begingroup$
Tank you for the explanation. Some questions that might sound stupid : If $n < omega$ what argument can we use to say that $3+n < omega$? also you say that $n<omega$ implies $n<omega$ and thus $3+omega leq omega$. So I guess the first statement should be $nleq omega$ implies $n leqomega$. This all may seem obvious to you but a lot of things that seemed obvious to me for normal numbers are not true anymore for ordinals.
$endgroup$
– roi_saumon
Jan 22 at 13:24
$begingroup$
@roi_saumon: I listed a lot of basic ordinal arithmetic results in this 29 October 2006 sci.math post, such as: "[$+$ is] associative, not commutative, continuous in the right variable, not continuous in the left variable $(sup{n+omega: n< omega }= omega< omega cdot 2 = sup{n: n < omega } + omega ),$ strictly increasing in the right variable, non-decreasing in the left variable (not strict: $1 + omega = 2 + omega).$" The proofs of these will depend on how your book/notes define ordinal addition, multiplication, etc.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:52
$begingroup$
@DaveL.Renfro I see. I am following notes that don't have proofs for this. Any good book with these proofs that you recommend?
$endgroup$
– roi_saumon
Jan 22 at 14:55
1
$begingroup$
@roi_saumon: Maybe Basic Set Theory by Shen/Vereshchagin, which doesn't have lengthy treatments of axiomatics and other issues that are rather tangential to someone only interested in ordinal arithmetic. Possibly the most detailed treatment of ordinal arithmetic I know of in a beginning set theory text is Introduction to Set Theory by James Donald Monk (1969), if you can find it in a library. And of course there is the bible.
$endgroup$
– Dave L. Renfro
Jan 22 at 15:31
$begingroup$
Tank you for the explanation. Some questions that might sound stupid : If $n < omega$ what argument can we use to say that $3+n < omega$? also you say that $n<omega$ implies $n<omega$ and thus $3+omega leq omega$. So I guess the first statement should be $nleq omega$ implies $n leqomega$. This all may seem obvious to you but a lot of things that seemed obvious to me for normal numbers are not true anymore for ordinals.
$endgroup$
– roi_saumon
Jan 22 at 13:24
$begingroup$
Tank you for the explanation. Some questions that might sound stupid : If $n < omega$ what argument can we use to say that $3+n < omega$? also you say that $n<omega$ implies $n<omega$ and thus $3+omega leq omega$. So I guess the first statement should be $nleq omega$ implies $n leqomega$. This all may seem obvious to you but a lot of things that seemed obvious to me for normal numbers are not true anymore for ordinals.
$endgroup$
– roi_saumon
Jan 22 at 13:24
$begingroup$
@roi_saumon: I listed a lot of basic ordinal arithmetic results in this 29 October 2006 sci.math post, such as: "[$+$ is] associative, not commutative, continuous in the right variable, not continuous in the left variable $(sup{n+omega: n< omega }= omega< omega cdot 2 = sup{n: n < omega } + omega ),$ strictly increasing in the right variable, non-decreasing in the left variable (not strict: $1 + omega = 2 + omega).$" The proofs of these will depend on how your book/notes define ordinal addition, multiplication, etc.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:52
$begingroup$
@roi_saumon: I listed a lot of basic ordinal arithmetic results in this 29 October 2006 sci.math post, such as: "[$+$ is] associative, not commutative, continuous in the right variable, not continuous in the left variable $(sup{n+omega: n< omega }= omega< omega cdot 2 = sup{n: n < omega } + omega ),$ strictly increasing in the right variable, non-decreasing in the left variable (not strict: $1 + omega = 2 + omega).$" The proofs of these will depend on how your book/notes define ordinal addition, multiplication, etc.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:52
$begingroup$
@DaveL.Renfro I see. I am following notes that don't have proofs for this. Any good book with these proofs that you recommend?
$endgroup$
– roi_saumon
Jan 22 at 14:55
$begingroup$
@DaveL.Renfro I see. I am following notes that don't have proofs for this. Any good book with these proofs that you recommend?
$endgroup$
– roi_saumon
Jan 22 at 14:55
1
1
$begingroup$
@roi_saumon: Maybe Basic Set Theory by Shen/Vereshchagin, which doesn't have lengthy treatments of axiomatics and other issues that are rather tangential to someone only interested in ordinal arithmetic. Possibly the most detailed treatment of ordinal arithmetic I know of in a beginning set theory text is Introduction to Set Theory by James Donald Monk (1969), if you can find it in a library. And of course there is the bible.
$endgroup$
– Dave L. Renfro
Jan 22 at 15:31
$begingroup$
@roi_saumon: Maybe Basic Set Theory by Shen/Vereshchagin, which doesn't have lengthy treatments of axiomatics and other issues that are rather tangential to someone only interested in ordinal arithmetic. Possibly the most detailed treatment of ordinal arithmetic I know of in a beginning set theory text is Introduction to Set Theory by James Donald Monk (1969), if you can find it in a library. And of course there is the bible.
$endgroup$
– Dave L. Renfro
Jan 22 at 15:31
add a comment |
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$begingroup$
Incidentally, "limit ordinal" is often defined to include $emptyset$, making $omega$ the second smallest.
$endgroup$
– J.G.
Jan 22 at 13:04